\(\int (1-\sec (e+f x))^m (d \sec (e+f x))^n \, dx\) [338]

Optimal result

Integrand size = 23, antiderivative size = 79 \[ \int (1-\sec (e+f x))^m (d \sec (e+f x))^n \, dx=-\frac {\operatorname {AppellF1}\left (n,\frac {1}{2}-m,\frac {1}{2},1+n,\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^n \tan (e+f x)}{f n \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \]

[Out]

-AppellF1(n,1/2,1/2-m,1+n,-sec(f*x+e),sec(f*x+e))*(d*sec(f*x+e))^n*tan(f*x+e)/f/n/(1-sec(f*x+e))^(1/2)/(1+sec(
f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3912, 138} \[ \int (1-\sec (e+f x))^m (d \sec (e+f x))^n \, dx=-\frac {\tan (e+f x) (d \sec (e+f x))^n \operatorname {AppellF1}\left (n,\frac {1}{2}-m,\frac {1}{2},n+1,\sec (e+f x),-\sec (e+f x)\right )}{f n \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}} \]

[In]

Int[(1 - Sec[e + f*x])^m*(d*Sec[e + f*x])^n,x]

[Out]

-((AppellF1[n, 1/2 - m, 1/2, 1 + n, Sec[e + f*x], -Sec[e + f*x]]*(d*Sec[e + f*x])^n*Tan[e + f*x])/(f*n*Sqrt[1
- Sec[e + f*x]]*Sqrt[1 + Sec[e + f*x]]))

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 3912

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^2*d
*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])), Subst[Int[(d*x)^(n - 1)*((a + b*x)^(m -
1/2)/Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !In
tegerQ[m] && GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {(d \tan (e+f x)) \text {Subst}\left (\int \frac {(1-x)^{-\frac {1}{2}+m} (d x)^{-1+n}}{\sqrt {1+x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \\ & = -\frac {\operatorname {AppellF1}\left (n,\frac {1}{2}-m,\frac {1}{2},1+n,\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^n \tan (e+f x)}{f n \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(257\) vs. \(2(79)=158\).

Time = 0.33 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.25 \[ \int (1-\sec (e+f x))^m (d \sec (e+f x))^n \, dx=\frac {(3+2 m) \operatorname {AppellF1}\left (\frac {1}{2}+m,m+n,1-n,\frac {3}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1-\sec (e+f x))^m (d \sec (e+f x))^n \sin (e+f x)}{f (1+2 m) \left ((3+2 m) \operatorname {AppellF1}\left (\frac {1}{2}+m,m+n,1-n,\frac {3}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+2 \left ((-1+n) \operatorname {AppellF1}\left (\frac {3}{2}+m,m+n,2-n,\frac {5}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+(m+n) \operatorname {AppellF1}\left (\frac {3}{2}+m,1+m+n,1-n,\frac {5}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[(1 - Sec[e + f*x])^m*(d*Sec[e + f*x])^n,x]

[Out]

((3 + 2*m)*AppellF1[1/2 + m, m + n, 1 - n, 3/2 + m, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 - Sec[e + f*x]
)^m*(d*Sec[e + f*x])^n*Sin[e + f*x])/(f*(1 + 2*m)*((3 + 2*m)*AppellF1[1/2 + m, m + n, 1 - n, 3/2 + m, Tan[(e +
f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF1[3/2 + m, m + n, 2 - n, 5/2 + m, Tan[(e + f*x)/2]^2, -T
an[(e + f*x)/2]^2] + (m + n)*AppellF1[3/2 + m, 1 + m + n, 1 - n, 5/2 + m, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2
]^2])*Tan[(e + f*x)/2]^2))

Maple [F]

\[\int \left (1-\sec \left (f x +e \right )\right )^{m} \left (d \sec \left (f x +e \right )\right )^{n}d x\]

[In]

int((1-sec(f*x+e))^m*(d*sec(f*x+e))^n,x)

[Out]

int((1-sec(f*x+e))^m*(d*sec(f*x+e))^n,x)

Fricas [F]

\[ \int (1-\sec (e+f x))^m (d \sec (e+f x))^n \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{n} {\left (-\sec \left (f x + e\right ) + 1\right )}^{m} \,d x } \]

[In]

integrate((1-sec(f*x+e))^m*(d*sec(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((d*sec(f*x + e))^n*(-sec(f*x + e) + 1)^m, x)

Sympy [F]

\[ \int (1-\sec (e+f x))^m (d \sec (e+f x))^n \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{n} \left (1 - \sec {\left (e + f x \right )}\right )^{m}\, dx \]

[In]

integrate((1-sec(f*x+e))**m*(d*sec(f*x+e))**n,x)

[Out]

Integral((d*sec(e + f*x))**n*(1 - sec(e + f*x))**m, x)

Maxima [F]

\[ \int (1-\sec (e+f x))^m (d \sec (e+f x))^n \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{n} {\left (-\sec \left (f x + e\right ) + 1\right )}^{m} \,d x } \]

[In]

integrate((1-sec(f*x+e))^m*(d*sec(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^n*(-sec(f*x + e) + 1)^m, x)

Giac [F]

\[ \int (1-\sec (e+f x))^m (d \sec (e+f x))^n \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{n} {\left (-\sec \left (f x + e\right ) + 1\right )}^{m} \,d x } \]

[In]

integrate((1-sec(f*x+e))^m*(d*sec(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^n*(-sec(f*x + e) + 1)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (1-\sec (e+f x))^m (d \sec (e+f x))^n \, dx=\int {\left (1-\frac {1}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

[In]

int((1 - 1/cos(e + f*x))^m*(d/cos(e + f*x))^n,x)

[Out]

int((1 - 1/cos(e + f*x))^m*(d/cos(e + f*x))^n, x)